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The Flamant solution provides expressions for the stresses and displacements in a linear elastic wedge loaded by point forces at its sharp end. This solution was developed by A. Flamant in 1892 by modifying the three-dimensional solution of Boussinesq.

The stresses predicted by the Flamant solution are (in polar coordinates)

- $$

- $$

(C_1costheta + C_3sintheta)~costheta~ dtheta = 0F_2 & + 2int_{alpha}^{beta}

(C_1costheta + C_3sintheta)~sintheta~ dtheta = 0end{align} where $F\_1,F\_2$ are the applied forces.

The wedge problem is self-similar and has no inherent length scale. Also, all quantities can be expressed in the separated-variable form $sigma\; =\; f(r)g(theta)$. The stresses vary as $(1/r)$.

For the special case where $alpha\; =\; -pi$, $beta\; =\; 0$, the wedge is converted into a half-plane with a normal force and a tangential force. In that case

- $$

- $$

- $$

- $begin\{align\}$

- $$

- $$

+1 & x > 0

-1 & x < 0end{cases}

- $$

- $begin\{align\}$

However, the concentrated loads at the vertex are difficult to express in terms of traction boundary conditions because

- the unit outward normal at the vertex is undefined
- the forces are applied at a point (which has zero area) and hence the traction at that point is infinite.

To get around this problem, we consider a bounded region of the wedge and consider equilibrium of the bounded wedge. Let the bounded wedge have two traction free surfaces and a third surface in the form of an arc of a circle with radius $a,$. Along the arc of the circle, the unit outward normal is $mathbf\{n\}\; =\; mathbf\{e\}\_r$ where the basis vectors are $(mathbf\{e\}\_r,\; mathbf\{e\}\_theta)$. The tractions on the arc are

- $$

Next, we examine the force and moment equilibrium in the bounded wedge and get

- $$

The traction-free boundary conditions on the edges $theta=alpha$ and $theta=beta$ also imply that

- $$

If we assume that $sigma\_\{rtheta\}=0$ everywhere, then the traction-free conditions and the moment equilibrium equation are satisfied and we are left with

- $$

~dtheta = 0F_2 & + int_{alpha}^{beta} sigma_{rr}(a,theta)~a~sintheta

~dtheta = 0end{align} and $sigma\_\{thetatheta\}\; =\; 0$ along $theta=alpha,\; theta=beta$ except at the point $r\; =\; 0$. But the field $sigma\_\{thetatheta\}\; =\; 0$ everywhere also satisfies the force equilibrium equations. Hence this must be the solution. Also, the assumption $sigma\_\{rtheta\}=0$ implies that $C\_2\; =\; C\_4\; =\; 0$.

Therefore,

- $$

To find a particular solution for $sigma\_\{rr\}$ we have to plug in the expression for $sigma\_\{rr\}$ into the force equilibrium equations to get a system of two equations which have to be solved for $C\_1,\; C\_3$:

- $$

(C_1costheta + C_3sintheta)~costheta~ dtheta = 0F_2 & + 2int_{alpha}^{beta}

(C_1costheta + C_3sintheta)~sintheta~ dtheta = 0end{align}

- $$

(C_1costheta + C_3sintheta)~costheta~ dtheta = 0 qquad implies F_1 + C_1pi = 0F_2 & + 2int_{-pi}^{0}

(C_1costheta + C_3sintheta)~sintheta~ dtheta = 0 qquad implies F_2 + C_3pi = 0end{align} Therefore

- $$

- $$

- $$

For $theta=0$ we have

- $$

- $$

- $$

- $$

- $$

- $$

+1 & x > 0

-1 & x < 0end{cases}

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This article is licensed under the GNU Free Documentation License.

Last updated on Wednesday October 01, 2008 at 19:40:48 PDT (GMT -0700)

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This article is licensed under the GNU Free Documentation License.

Last updated on Wednesday October 01, 2008 at 19:40:48 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

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